Question

10. Camider the ring of plynicanials z,Ir, and let/ denote the elmmont r4 + 2a + 1 a) (5 points) Show that the quotient rga)/ () is a field. b) (5 points) Let a denote the coset z()Regarding F as a vector space over Z2, find a basis for F coasisting of powers of a c) (5 poluts) How nuany elements dors F have? Justify your answer. d) (5 points) Compute the product afas t a) i.e. expand this product on the basis you found above). e) (5 points) Find the maitiplicative iaverse of a f0) (5 points) The polynonial f factors over F into a product of linear factors. Find this factorization. (Hint: gou need to find the roots of f in F. This can certainty be done by brute force, though there is a faster way: it root ( bertius niap" o(t)-ra is a symmetry of the crtension Z2 → F. Since a is , it fotlous that ofa), olela)). etc. are also roots of s) (5 points) Suppose f e Ri is ireducible. Prove that degif) s2. (Hint: the field R/) is an algebraic extension of R, and thus can be embedded the ะ Finally, yosreld then Imust be divisible orguing that j must have some comensons, Altern atítely, you could shone that if z is not real, then f must be divisible by the t nto the algebraic closure of R. Now count dimensions. re or less euivalently, you could begin by arguing that f could show that if z is real, then f ust be divisible by the real must have some complex root s. Then -2 and must hence be real polmomial (r-2) b oft and must hence be an associate of (ax- r 2). But the mension-counting aryument is faster.) h zu Qwetio:Let r) (5 bnu s points) Suppoeie f EZplr) is irreducible, putdedeg(J), and set puta des) onut: the structure F-7,1 r / (J) is a field uith p d elements. As usual, put α the mininal polynonial of a over Zp. What is g(a)?) In this question you will develop a fst irreducibili Bonus Ques tion: Let p be a prime. will develop a fast irreducibility test for polynomials you g (r")-r. Show that f Then f is elements. As usual, put a+(). Then f is with p + ) Let f and d be as above. Show that f does not divide z(r)-z for any integer i with b) (5 bonus points 0isd. (Hint: first show that if it did, then I would splt oter GF(p). Then the subfield of GF(p) generated ra r f eed Pinali dre onudiction by dentifyng the true sphliting field of f.) c) (5 bonus points) Now let JE Z,lar] be auy non-constant polyaouial. Show that f is irreducible if and ouly if ged(f)x) 1 for every i with 0 < i < degíf). (Thus, if one uses the Euclidean algorithun to compute greatest common divisors, then one obtains an algorithmic irreducibility test whose running tine is polynomial in degif). At first glance this algorithm seems to have exponential time and memory consumption, since one naust apparently divide the polynomial lo)-z by f, but this can be avoided if the first division in the Euclidean algorithm is replaced by a clever trick: one computes o (mod f) by repeated squaring, then subtracts z to obtain the first remamder, and then one completes the Euclidean algorithn as usual. For a slightly faster test, look up Rabin's test for irreducibility.)

Answer 1

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Observe that Z_2[x] is a PID & f(x) = x^4 + x^3 + 1 is irreducible over Z_2 since it has only one possible factorization (x^2 - 1) (x^2 + 1) notequalto x^4 + x^3 + 1 (f(x) has no roots in Z_2) thus f(x) is prime element in Z_2 [x] thus < f > is prime element in Z_2 [x] thus < f > is prime ideal = > maximal ideal in Z_2 [x] F = Z_2 [x]/ < f > is field. Observe that 1, alpha, alpha^2 alpha^3 are linearly independent over F ie alpha^4 = x^4 + < f > but x^4 + x^3 + 1 = 0 = > x^4 = -x^3 - 1 so x^4 is dependent on alpha^3, & 1. So {1, alpha, alpha^2, alpha^3} is a basis for F. For any element of F is if the form a + b alpha + c alpha^2 + d alpha^3 each a, b, c, d elementof Z_2 so total (2)^4 = 16 possibility